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faust18 [17]
3 years ago
14

Jeff used an equal arm balance to weigh a 4.312 g sample of sodium chloride. Which of these measurements made by Jeff is the mos

t accurate? (5 points) 4.5 g 4.1 g 4.2 g 4.3 g
Chemistry
2 answers:
AlladinOne [14]3 years ago
4 0

<u>Answer:</u> The most accurate measurement made by Jeff is 4.3 g

<u>Explanation:</u>

Accuracy is defined as the closeness of a measured value to a standard or known value.

For Example: If the mass of a substance is 58 kg and one person weighed 57 kg and another person weighed 60 kg. Then, the weight measured by first person is more accurate.

We are given:

Mass of the sample = 4.312 g

So, the measurement close to 4.312 grams will be the accurate value.

Hence, the most accurate measurement made by Jeff is 4.3 g

Virty [35]3 years ago
3 0

Answer:

4.3.

Explanation:

  • Measurements that are close to the known value are said to be accurate, whereas measurements that are close to each other are said to be precise.
  • The most closest measurement to the known value is 4.3 g.
  • So, the right choice is: 4.3 g.
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A 0.8115 g sample of HCl was placed into a 50 mL volumetric flask and the sample was thoroughly dissolved in water to make 50 mL
inysia [295]

Answer:

Molarity of NaOH solution is 1.009 M

Explanation:

Molar mass of HCl is 36.46 g/mol

Number moles = (mass)/(molar mass)

So, 0.8115 g of HCl = \frac{0.8115}{36.46}moles HCl = 0.02226 moles HCl

1 mol of NaOH neutralizes 1 mol of HCl.

So, if molarity of NaOH solution is S(M) then moles of NaOH required to reach endpoint is \frac{S\times 22.07}{1000}moles

So, \frac{S\times 22.07}{1000}=0.02226

or, S = 1.009

So, molarity of NaOH solution is 1.009 M

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Stolb23 [73]
Mass of Lead is .00042 g
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The metalloid that has three valence electrons is .
lawyer [7]

Answer:

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Explanation:

4 0
2 years ago
Read 2 more answers
A 20.0 mL 0.100 M solution of lactic acid is titrated with 0.100 M NaOH.
yan [13]

Answer:

(a) See explanation below

(b) 0.002 mol

(c) (i) pH = 2.4

(ii) pH = 3.4

(iii) pH = 3.9

(iv) pH = 8.3

(v) pH = 12.0

Explanation:

(a) A buffer solution exits after addition of 5 mL of NaOH  since after reaction we will have  both the conjugate base lactate anion and unreacted weak  lactic acid present in solution.

Lets call lactic acid HA, and A⁻ the lactate conjugate base. The reaction is:

HA + NaOH ⇒ A⁻ + H₂O

Some unreacted HA will remain in solution, and since HA is a weak acid , we will have the followin equilibrium:

HA  + H₂O ⇆ H₃O⁺ + A⁻

Since we are going to have unreacted acid, and some conjugate base, the buffer has the capacity of maintaining the pH in a narrow range if we add acid or base within certain limits.

An added acid will be consumed by the conjugate base A⁻ , thus keeping the pH more or less equal:

A⁻ + H⁺ ⇄ HA

On the contrary, if we add extra base it will be consumed by the unreacted lactic acid, again maintaining the pH more or less constant.

H₃O⁺ + B ⇆ BH⁺

b) Again letting HA stand for lactic acid:

mol HA =  (20.0 mL x  1 L/1000 mL) x 0.100 mol/L = 0.002 mol

c)

i) After 0.00 mL of NaOH have been added

In this case we just have to determine the pH of a weak acid, and we know for a monopric acid:

pH = - log [H₃O⁺] where  [H₃O⁺] = √( Ka [HA])

Ka for lactic acid = 1.4 x 10⁻⁴  ( from reference tables)

[H₃O⁺] = √( Ka [HA]) = √(1.4 x 10⁻⁴ x 0.100) = 3.7 x 10⁻³

pH = - log(3.7 x 10⁻³) = 2.4

ii) After 5.00 mL of NaOH have been added ( 5x 10⁻³ L x 0.1 = 0.005 mol NaOH)

Now we have a buffer solution and must use the Henderson-Hasselbach equation.

                            HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn         0.002                  0.0005                0

after rxn    0.002-0.0005                  0                  0.0005

                        0.0015

Using Henderson-Hasselbach equation :

pH = pKa + log [A⁻]/[HA]

pKa HA = -log (1.4 x 10⁻⁴) = 3.85

pH = 3.85 + log(0.0005/0.0015)

pH = 3.4

iii) After 10.0 mL of NaOH have been ( 0.010 L x 0.1 mol/L = 0.001 mol)

                             HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn         0.002                  0.001               0

after rxn        0.002-0.001                  0                  0.001

                        0.001

pH = 3.85 + log(0.001/0.001)  = 3.85

iv) After 20.0 mL of NaOH have been added ( 0.002 mol )

                            HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn         0.002                  0.002                 0

after rxn                 0                         0                   0.002

We are at the neutralization point and  we do not have a buffer anymore, instead we just have  a weak base A⁻ to which we can determine its pOH as follows:

pOH = √Kb x [A⁻]

We need to determine the concentration of the weak base which is the mol per volume in liters.

At this stage of the titration we added 20 mL of lactic acid and 20 mL of NaOH, hence the volume of solution is 40 mL (0.04 L).

The molarity of A⁻ is then

[A⁻] = 0.002 mol / 0.04 L = 0.05 M

Kb is equal to

Ka x Kb = Kw ⇒ Kb = 10⁻¹⁴/ 1.4 x 10⁻⁴ = 7.1 x 10⁻¹¹

pOH is then:

[OH⁻] = √Kb x [A⁻]  = √( 7.1 x 10⁻¹¹ x 0.05) = 1.88 x 10⁻⁶

pOH = - log (  1.88 x 10⁻⁶ ) = 5.7

pH = 14 - pOH = 14 - 5.7 = 8.3

v) After 25.0 mL of NaOH have been added (

                            HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn           0.002                  0.0025              0

after rxn                0                         0.0005              0.0005

Now here what we have is  the strong base sodium hydroxide and A⁻ but the strong base NaOH will predominate and drive the pH over the weak base A⁻.

So we treat this part as the determination of the pH of a strong base.

V= (20 mL + 25 mL) x 1 L /1000 mL = 0.045 L

[OH⁻] = 0.0005 mol / 0.045 L = 0.011 M

pOH = - log (0.011) = 2

pH = 14 - 1.95 = 12

7 0
2 years ago
What do chemical and physical change have in common?
gtnhenbr [62]

Explanation:

The similarities between chemical and physical changes include change in state of matter and both process involve change in energy.

Chemical changes of a substance results in the formation of entirely new substances. These changes are not reversible.

Physical changes results in the change of the state of the substance which can be reversed.

However, both chemical and physical changes results in change of state of matter and both process results in change in energy.

Thus, we can conclude that the similarities between chemical and physical changes include change in state of matter and both process involve change in energy.

(it's alre answered but here it again 'NOTE: not my answer')

3 0
1 year ago
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